The produce manager for a large retail food chain is interested in estimating the percentage of potatoes that arrive on a shipment with bruises. A random sample of 150 potatoes showed 14 with bruises. Based on this information, what is the margin of error for a 95 percent confidence interval estimate?
A) 0.0933
B) 0.0466
C) 0.0006
D) Can't be determined without knowing σ.

Question

Quizplus · Accepted Answer

The margin of error for a proportion is calculated using the formula $ME = Z \times \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score for the confidence level (1.96 for 95%), $p$ is the sample proportion, and $n$ is the sample size. Here, $p = \frac{14}{150}$ and $n = 150$. Plugging these values in gives $ME = 1.96 \times \sqrt{\frac{(14/150)(1 - 14/150)}{150}} \approx 0.0466$.

The Produce Manager for a Large Retail Food Chain Is